Tuned Power Lines

Demonstrative Video

  • Wavelength, \(\lambda\): distance along the line between two points of a wave which differ in phase by \(360^0\) or \(2\pi\) radian.

  • \[\lambda = \dfrac{2\pi}{\beta}\]
    is the phase shift, then If
  • \[\mbox{velocity} = \dfrac{\lambda}{t}= \lambda \cdot f = \dfrac{2\pi f}{\beta}\]
    The velocity of propagation of wave
  • \[\begin{array}{ccccc} \lambda= & \dfrac{1}{f\sqrt{LC}} & & & \mbox{velocity}=\dfrac{1}{\sqrt{LC}}\end{array}\]
    For the lossless line and
  • \[\lambda \simeq \dfrac{3 \times 10^8}{50} = 6000 ~ \mbox{Km}\]
    Wavelength of a 50 Hz power transmission
\[\begin{aligned} \gamma & =\sqrt{yz} =\sqrt{j\omega C\left(r+j\omega L\right)}\\ \alpha+j\beta & =j\omega\sqrt{LC}\left(1-j\dfrac{r}{\omega L}\right)^{1/2}\\ & =j\omega\sqrt{LC}\left(1-j\dfrac{r}{2\omega L}\right) \end{aligned}\]
\[\begin{aligned} L & =\dfrac{\mu_{0}}{2\pi}ln\dfrac{D}{r^{'}}\\ C & =\dfrac{2\pi\varepsilon_{0}}{ln\left(D/r\right)}\\ v & =\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}\\ & =\mbox{velocity of light} \end{aligned}\]

Tuned Power Lines

\[\begin{aligned} \gamma & =\sqrt{yz}=j\omega\sqrt{LC}\\ \Longrightarrow cosh\left(\gamma l\right) & =cosh\left(j\omega l\sqrt{LC}\right)=cos\left(\omega l\sqrt{LC}\right)\\ \Longrightarrow sinh\left(\gamma l\right) & =sinh\left(j\omega l\sqrt{LC}\right)=jsin\left(\omega l\sqrt{LC}\right) \end{aligned}\]
\[\left[\begin{array}{c} V_{S}\\ I_{S} \end{array}\right]=\left[\begin{array}{cc} cos\left(\omega l\sqrt{LC}\right) & jZ_{c}sin\left(\omega l\sqrt{LC}\right)\\ \dfrac{j}{Z_{c}}sin\left(\omega l\sqrt{LC}\right) & cos\left(\omega l\sqrt{LC}\right) \end{array}\right]\left[\begin{array}{c} V_{R}\\ I_{R} \end{array}\right]\]
\[\begin{aligned} |V_{S}| & =|V_{R}|\\ |I_{S}| & =|I_{R}| \end{aligned}\]
\(\omega l\sqrt{LC} =n\pi,~n=1,2,3\cdots\)On simplification,

  • the receiving-end voltage and current are numerically equal to the corresponding sending-end values,

  • there is no voltage drop on load

  • Such a line is called a .

\[l=\dfrac{n\pi}{2\pi f\sqrt{LC}}\]
\[\begin{aligned} l & =\dfrac{1}{2}\left(n\lambda\right)=\dfrac{1}{2}\lambda,\lambda,\dfrac{3}{2}\lambda,\cdots\\ & =3000~km,~6000~km,\cdots \end{aligned}\]
\(1/\sqrt{LC}\simeq\vartheta,\)For 50 Hz, the length of line for tuning is

  • too long a distance of transmission from the point of view of cost and efficiency

  • for a given line, length and frequency tuning can be achieved by increasing \(L\) or \(C\), i.e. by adding series inductances or shunt capacitances at several places along the line length

  • method is impractical and uneconomical for power frequency lines and is adopted for telephony where higher frequencies are employed

  • practical method employs series capacitor to reduce the line inductance and shunt inductor to neutralize line capacitance